Question

A cubical distribution of charge has a uniform volume charge density of 6√3 x 10 Cm3. The total charge in the distribution is 20 mC. If the same charge is uniformly distributed over the entire volume of a sphere of diameter equal to the leading diagonal of the cube, then the new volume charge density will be:​

Answer 1

Answer:

A cubical distribution of charge has a uniform volume charge density of 6√3 x 10 Cm3. The total charge in the distribution is 20 mC. If the same charge is uniformly distributed over the entire volume of a sphere of diameter equal to the leading diagonal of the cube, then the new volume charge density will be $0.606 C/m^{3}$

Explanation:

• The volume charge density (ρ) is the charge (q) in unit volume (v), that is

        ρ= $q/v$

• For a cube with side length (x) volume is given by $V=x^{3}$ and the length of the leading diagonal is $\sqrt{3}x$

• For a sphere of radius (r) volume is given by $v=\frac{4}{3} \pi r^{3}$

From the question, we have

the volume charge density of the cube ρ=$6\sqrt{3}*10C/m^{3}$

the total charge present $q=20*10^{-6}C$

now the volume of the cube $v=q/$ρ

⇒$v=\frac{20*10^{-6} }{6\sqrt{3}*10 } =1.9*10^{-5}$

so one side of the cube $x=0.027m$

the diameter is equal to the leading diagonal length

the leading diagonal length $l=\sqrt{3}*0.027 =0.049m$

so the radius of the sphere $r=0.02m$

the volume of the sphere $v=\frac{4}{3}* \pi* 0.02^{3}=3.3*10^{-5}m^{3}$

since the same charge is distributed the new volume charge density is

ρ= $\frac{20*10^{-6} }{3.3*10^{-5} } =0.606 C/m^{3}$